MA1100 Tutorial Xiang Sun 12 Dept. Mathematics November 15, 2009 - PowerPoint PPT Presentation

. Tutorial 1: Sets and Logic . Solution. Exercise (2.13) Tutorial contradiction, there is no such A . MA1100 Tutorial . . . . Let S = { 1 , 2 , . . . , 6 } and let P ( A ) : A X { 2 , 4 , 6 } = H and Q ( A ) : A � H be open sentence over the domain P ( S ) . (a) Determine all A P P ( S ) for which P ( A ) ^ Q ( A ) is true. (b) Determine all A P P ( S ) for which P ( A ) _ ( � Q ( A )) is true. (c) Determine all A P P ( S ) for which ( � P ( A )) ^ ( � Q ( A )) is true. { P ( A ) is true � A X { 2 , 4 , 6 } = H (a) P ( A ) ^ Q ( A ) is true � (and) � A must Q ( A ) is true � A � H be a non-empty subset of { 1 , 3 , 5 } ; { P ( A ) is true � A X { 2 , 4 , 6 } = H (b) P ( A ) _ ( � Q ( A )) is true � (or) � A Q ( A ) is false � A = H must be a subset of { 1 , 3 , 5 } . { P ( A ) is false � A X { 2 , 4 , 6 } � H (c) ( � P ( A )) ^ ( � Q ( A )) is true � (and) � Q ( A ) is false � A = H

. Solution. . . . . MA1100 Tutorial Tutorial 1: Sets and Logic Tutorial Exercise (2.21) . given values of x and y . In each of the following, two open sentences P ( x , y ) and Q ( x , y ) are given, where the domain of both x and y is Z . Determine the truth value of P ( x , y ) � Q ( x , y ) for the (a) P ( x , y ) : x 2 � y 2 = 0 and Q ( x , y ) : x = y . ( x , y ) P { (1 , � 1) , (3 , 4) , (5 , 5) } . (b) P ( x , y ) : | x | = | y | and Q ( x , y ) : x = y . ( x , y ) P { (1 , 2) , (2 , � 2) , (6 , 6) } . (c) P ( x , y ) : x 2 + y 2 = 1 and Q ( x , y ) : x + y = 1 . ( x , y ) P { (1 , � 1) , ( � 3 , 4) , (0 , � 1) , (1 , 0) } . (a) When ( x , y ) = (1 , � 1) , P ( x , y ) is true, Q ( x , y ) is false, then P ( x , y ) � Q ( x , y ) is false; Similarly, truth for ( x , y ) = (3 , 4) and ( x , y ) = (5 , 5) ; (b) Truth for ( x , y ) = (1 , 2) and ( x , y ) = (6 , 6) , false for ( x , y ) = (2 , � 2) ; (c) Truth for ( x , y ) = (1 , � 1) , ( x , y ) = ( � 3 , 4) and ( x , y ) = (1 , 0) , false for ( x , y ) = (0 , � 1) .

. . . . . . MA1100 Tutorial Tutorial 1: Sets and Logic Additional material Russell’s paradox Exercise Usually, for any formal criterion, a set exists whose members are those objects (and only those objects) that satisfy the criterion, i.e. { x P U : p ( x ) } is a set. Whether does there exist an object with the form { x P U : p ( x ) } , which is not a set?

. Russell’s paradox the same definition. This contradiction is Russell’s paradox. hand, if such a set is not a member of itself, it would qualify as a member of itself by definition as a set containing sets that are not members of themselves. On the other themselves. If such a set qualifies as a member of itself, it would contradict its own This question is disproved by a set containing exactly the sets that are not members of . Solution. Additional material Tutorial 1: Sets and Logic MA1100 Tutorial . . . . set). Let A = { X P U : X R X } , U is the collection of all sets. If A P A , then A does not satisfy X R X , i.e. A R A , contradiction; If A R A , then A satisfies X R X , i.e. A P A , contradiction. Therefore, the definition is not well-defined, and the error is from U ( � U is not a

. Solution. now-canonical Zermelo-Fraenkel set theory (ZF). axioms of extensionality and unlimited set abstraction, and evolved into the constructed axiomatic set theory. Zermelo’s axioms went well beyond Frege’s Russell’s type theory and Ernst Zermelo’s axiomatic set theory, the first Russell in 1901. In 1908, two ways of avoiding the paradox were proposed, Russell’s paradox (also known as Russell’s antinomy), discovered by Bertrand Roughly speaking, the method is giving some restrictions on the definition of set. How to solve this problem? . Exercise Additional material Tutorial 1: Sets and Logic MA1100 Tutorial . . . . For more information, you can wiki “Russell’s paradox”.

. . . . . . MA1100 Tutorial Tutorial 2: Logic Schedule of Today Review concepts Tutorial: 2.31(contradiction), 2.32(tautology), 2.37(equivalence), 2.40(negation), 2.48, 2.49(negation), 2.62(equivalence), 2.67, 2.68

. Distributive Law Logical Equivalence Two logical expressions are said to be logically equivalent to each Contradiction A logical expression that is always false is called a contradiction. Tautology A logical expression that is always true is called a tautology. . Logic II(lecture 5) Review Tutorial 2: Logic MA1100 Tutorial . . . . other if they have the same truth value. Biconditional P if and only if Q , that is P � Q . Converse Q � P is called the converse of P � Q . Contrapositive ( � Q ) � ( � P ) is called the contrapositive of P � Q . De Morgan’s Law � ( P ^ Q ) � ( � P ) _ ( � Q ) , � ( P _ Q ) � ( � P ) ^ ( � Q ) . P _ ( Q ^ R ) � ( P _ Q ) ^ ( P _ R ) , P ^ ( Q _ R ) � ( P ^ Q ) _ ( P ^ R ) . Implication as disjunction P � Q � ( � P ) _ Q . Negation of implication � ( P � Q ) � P ^ ( � Q ) . Implication with disjunction P � ( Q _ R ) � ( P ^ ( � Q )) � R .

. False True False only some x True True all the x . False Existential quantifier The phrase “there exists”, “there is”, . . . is called a existential Review Logic III(lecture 6) Tutorial 2: Logic MA1100 Tutorial . . . . none of the x Universal quantifier The phrase “for each”, “for every”, “for all”, . . . is called a universal quantifier. Notation: @ , say “for all”. quantifier. Notation: D , say “there exist”. P ( x ) true for ( @ x ) P ( x ) ( D x ) P ( x ) @ vs D Two quantifiers ( @ x )( @ y ) P ( x , y ) , ( D x )( D y ) P ( x , y ) , ( @ y )( D x ) P ( x , y ) , ( D x )( @ y ) P ( x , y ) . Negation with quantifier � ( @ x ) P ( x ) � ( D x )( � P ( x )) , � ( D x ) P ( x ) � ( @ x )( � P ( x )) , � ( @ x )( D y ) P ( x , y ) � ( D x )( @ y )( � P ( x , y )) , � ( D x )( @ y ) P ( x , y ) � ( @ x )( D y )( � P ( x , y )) , � ( @ x )( @ y ) P ( x , y ) � ( D x )( D y )( � P ( x , y )) , � ( D x )( D y ) P ( x , y ) � ( @ x )( @ y )( � P ( x , y ))

. T T T F T F F T F T F F Q F T F F F F F F T F F . F P General method: It suffices to show that the statement is false for all combinations. . . . . MA1100 Tutorial Tutorial 2: Logic Tutorial Exercise (2.31) For statements P and Q , show that Recall A logical expression that is always false is called a contradiction. How to prove that a truth table below. statement is contradiction? Proof of Method 1. for all combinations of truth values for the component statements P and Q . See the The compound statement ( ) P ^ ( � Q ) ^ ( P ^ Q ) is a contradiction. ( ) P ^ ( � Q ) ^ ( P ^ Q ) is a contradiction since it is false ( ) � Q P ^ Q P ^ ( � Q ) P ^ ( � Q ) ^ ( P ^ Q )

. 2 i.e. . 4 . . . By commutative law, we have 1 . . . . 3 Also by associated law, we have By associated law, we have . . Exercise (2.31) . . . . MA1100 Tutorial Tutorial 2: Logic Tutorial For statements P and Q , show that . Proof of Method 2. . . ( ) P ^ ( � Q ) ^ ( P ^ Q ) is a contradiction. ( ) P ^ ( � Q ) ^ ( P ^ Q ) � P ^ ( � Q ) ^ P ^ Q . P ^ ( � Q ) ^ P ^ Q � P ^ P ^ ( � Q ) ^ Q . P ^ P ^ ( � Q ) ^ Q � ( P ^ P ) ^ ( Q ^ ( � Q )) � P ^ ( Q ^ ( � Q )) . Since Q ^ ( � Q ) is always false, we have that P ^ ( Q ^ ( � Q )) is contradiction, ( ) P ^ ( � Q ) ^ ( P ^ Q ) is a contradiction.

. F . Q T T T T T T F F T combinations of truth values for the component statements P and Q . See the truth F T T F T F F T F T table below. P Tutorial For statements P and Q , show that modus ponens.) Recall A logical expression that is always true is called a tautology. How to prove that a Exercise (2.32) Tutorial 2: Logic MA1100 Tutorial . . statement is tautology? General method: It suffices to show that the statement is true for all combinations. Proof of Method 1. The compound statement . . ( ) P ^ ( P � Q ) � Q is a tautology. Then state ( ) P ^ ( P � Q ) � Q in words. (This is an important logical argument form, called ( ) P ^ ( P � Q ) � Q is a tautology since it is true for all ( ) P � Q P ^ ( P � Q ) P ^ ( P � Q ) � Q ( ) P ^ ( P � Q ) � Q : If P and P implies Q , then Q .

. 3 . . . 2 contradiction, i.e., this case will not happen, then it is done. . . . Suppose that R is true and Q is false. . . . . 4 . . . 5 We get that Q is true. It is a contradiction. Therefore, this case will not . 1 . . . . . . MA1100 Tutorial Tutorial 2: Logic Tutorial Exercise (2.32) For statements P and Q , show that happen, that is, the statement is a tautology. Proof of Method 2. modus ponens.) ( ) P ^ ( P � Q ) � Q is a tautology. Then state ( ) P ^ ( P � Q ) � Q in words. (This is an important logical argument form, called Let R = P ^ ( P � Q ) , we want to show R � Q is always true, that is, the cases in which R � Q is false will not happen. R � Q is false only when R is true and Q is false. If we prove that there is a Since R = P ^ ( P � Q ) , we have that P and P � Q are both true.

. F T F F F T T F F T F F . T T F F T F F T T F T F Proof of Method 1. . Tutorial 2: Logic Tutorial Exercise (2.37) equivalent. Recall Two logical statements are equivalent if they have the same truth value. How to prove that two statements are equivalent? General method: It suffices to show that two statements have same truth value. . MA1100 Tutorial are logically equivalent since they have the same truth values for all combinations of truth values for the component statements P and Q . See the truth table below. P . . Q For statements P and Q , show that ( � Q ) � ( P ^ ( � P )) and Q are logically ( ) The statements Q and ( � Q ) � P ^ ( � P ) ( ) � P � Q P ^ ( � P ) ( � Q ) � P ^ ( � P )

. . 3 . . . 2 . . . 1 . . . Proof of Method 2. equivalent. Exercise (2.37) Tutorial Tutorial 2: Logic MA1100 Tutorial . . . . For statements P and Q , show that ( � Q ) � ( P ^ ( � P )) and Q are logically Since P ^ ( � P ) is always false, it is enough to consider Q . When Q is true, then ( � Q ) � ( P ^ ( � P )) is true, the same as Q . When Q is false, then ( � Q ) � ( P ^ ( � P )) is false, the same as Q .

. Solution. . . . . MA1100 Tutorial Tutorial 2: Logic Tutorial Exercise (2.40) Write negations of the following open sentences: . (b) The integers a and b are both even. Recall (a) Either x = 0 or y = 0 . De Morgan’s laws: � ( P ^ Q ) � ( � P ) _ ( � Q ) , � ( P _ Q ) � ( � P ) ^ ( � Q ) . Roughly speaking, we can consider � as an operator, which changes ^ (and) to _ (or), and changes _ (or) to ^ (and). (a) Let P : x = 0 and Q : y = 0 . Then the statement can be expressed as P _ Q . Therefore the negation is ( � P ) ^ ( � Q ) , i.e. both x � 0 and y � 0 . (b) Let P : a is even and Q : b is even, then the statement can be expressed as P ^ Q . Therefore the negation is ( � P ) _ ( � Q ) , i.e. either a is odd or b is odd.

. Tutorial False True . Determine the truth value of each of the following statements. Exercise (2.48) Recall Tutorial 2: Logic . . MA1100 Tutorial . . (a) D x P R , x 2 � x = 0 . (f) @ x , y P R , x + y + 3 = 8 . (b) @ n P N , n + 1 � 2 . (g) D x , y P R , x 2 + y 2 = 9 . √ x 2 = x . (c) @ x P R , (h) @ x P R , @ y P R , x 2 + y 2 = 9 . (d) D x P Q , 3 x 2 � 27 = 0 . (e) D x P R , D y P R , x + y + 3 = 8 . ( @ x ) P ( x ) P ( x ) is true for all x P ( x ) is false for some x ( D x ) P ( x ) P ( x ) is true for some x P ( x ) is false for all x ( @ x )( @ y ) P ( x , y ) P ( x , y ) is true for all x and all y P ( x , y ) is false for some x or some y ( D x )( D y ) P ( x , y ) P ( x , y ) is true for some x and some y P ( x , y ) is false for all x and y ( D x )( @ y ) P ( x , y ) For some x , P ( x , y ) is true for all y For any x , P ( x , y ) is false for some y ( @ x )( D y ) P ( x , y ) For any x , P ( x , y ) is true for some y For some x , P ( x , y ) is false for all y

. . . . . . MA1100 Tutorial Tutorial 2: Logic Tutorial Solution. (a) True. 0 and 1 are the solutions of x 2 � x = 0 . (b) True. For any n P N , we have n � 1 since N = { 1 , 2 , 3 , . . . } . Therefore n + 1 � 2 . √ x 2 = � x � x . (c) False. Let x � 0 , then (d) True. 3 and -3 are the solutions of 3 x 2 = 27 . (e) True. x = 0 , y = 5 is the solution of x + y + 3 = 8 . (f) False. x = 0 , y = 0 does not satisfy x + y + 3 = 8 . (g) True. x = 0 , y = 3 is the solution of x 2 + y 2 = 9 . (h) False. x = 0 , y = 0 does not satisfy x 2 + y 2 = 9 .

. Exercise (2.49) (d) Using quantifiers, express in symbols the negations of the statements in both (a) (c) Express in words the negations of the statements in (a) and (b). Do this for the statements in parts (a) and (b). can be expressed using a quantifier as: . The statement Tutorial Tutorial 2: Logic MA1100 Tutorial . . . . and (b). For every integer m , either m � 1 or m 2 � 4 . @ m P Z , m � 1 or m 2 � 4 . (a) There exist integers a and b such that both ab � 0 and a + b � 0 . (b) For all real numbers x and y , x � y implies that x 2 + y 2 � 0 .

. Solution. . . . . MA1100 Tutorial Tutorial 2: Logic Tutorial Recall . (c-d) Roughly speaking, � changes @ to D , and changes D to @ . Implication as disjunction: P � Q � ( � P ) _ Q . Negation of implication: � ( P � Q ) � P ^ ( � Q ) . (a) D a , b P Z , ab � 0 and a + b � 0 . (b) @ x , y P R , x � y implies x 2 + y 2 � 0 . (a) @ a , b P Z , either ab � 0 or a + b � 0 , that is, for all integers a and b , either ab � 0 or a + b � 0 . (b) D x , y P R , x � y and x 2 + y 2 � 0 , that is, there exist real numbers x and y such that x � y and x 2 + y 2 � 0 . Besides, the negation of (b) can be: D x , y P R , x � y and x 2 + y 2 = 0 since x 2 + y 2 � 0 for all x , y P R .

. . . . . MA1100 Tutorial Tutorial 2: Logic Tutorial Exercise (2.62) . (a) For statements P , Q , and R , show that ( ) ( ) ( P ^ Q ) � R � ( P ^ ( � R )) � ( � Q ) . (b) For statements P , Q , and R , show that ( ) ( ) ( P ^ Q ) � R � ( Q ^ ( � R )) � ( � P ) .

. T F T F F F F T F T T T T F F F T T T T F Part(b): F F . T F F T T P F Q R T T T F F T F T T T T F T F T T F T T F F F F T T F F T T F T F F F F T T T T F T T T T F F T F F F F T T T T T F T T F F F T F F T T F F F F T T T F The logical expressions are said to be locally equivalent to each other if they have the . . . . MA1100 Tutorial Tutorial 2: Logic Tutorial Proof of Method 1. same truth value. F Part(a): P Q R T T T F T F F F F T F F T F T T T F T T F T T T F F T F F T T T T T T F F F T ( ) ( ) � Q � R P ^ Q P ^ ( � R ) ( P ^ Q ) � R ( P ^ ( � R )) � ( � Q ) Q ^ ( � R ) ( ( P ^ Q ) � R ) ( ( Q ^ ( � R )) � ( � P ) ) � P � R P ^ Q

. . (b) . . . 1 . . . Therefore, we have 2 By commutative law, we have . . . . 3 . . by commutative law by De Morgan’s law . . . . MA1100 Tutorial Tutorial 2: Logic Tutorial Proof of Method 2. (a) by implication as disjunction Apply part (a). by De Morgan’s law by implication as disjunction ( P ^ Q ) � R �� ( P ^ Q ) _ R � ( � P ) _ ( � Q ) _ R ( P ^ ( � R )) � ( � Q ) �� ( P ^ ( � R )) _ ( � Q ) � ( � P ) _ R _ ( � Q ) � ( � P ) _ ( � Q ) _ R ( ) ( ) ( P ^ Q ) � R � ( P ^ ( � R )) � ( � Q ) Let Q 1 = P and P 1 = Q , then the statement can be written as ( ( Q 1 ^ P 1 ) � R ) ( ( P 1 ^ ( � R )) � ( � Q 1 ) ) � ( ( P 1 ^ Q 1 ) � R ) ( ( P 1 ^ ( � R )) � ( � Q 1 ) ) �

. . 2 . . . 3 . . 4 . . . . 5 . . . 6 . . . Exercise (2.67) . . . . MA1100 Tutorial Tutorial 2: Logic Tutorial Therefore, there does not exist such set S . Solution. . . . 1 Do there exist a set S of cardinality 2 and a set { P ( n ) , Q ( n ) , R ( n ) } of three open sentences over the domain S such that the implications P ( a ) � Q ( a ) , Q ( b ) � R ( b ) , and R ( c ) � P ( c ) are true, where a , b , c P S , and (2) the converses of the implications in (1) are false? Necessarily, at least two of these elements a , b . If Q ( a ) � P ( a ) , R ( b ) � Q ( b ) , and P ( c ) � R ( c ) are false, then Q ( a ) , R ( b ) , P ( c ) must be true, and P ( a ) , Q ( b ) , R ( c ) must be false. Since | S | = 2 , there are at least two of a , b , c which are same. If a = b , since Q ( a ) is true and Q ( b ) is false, it is a contradiction. If a = c , since P ( c ) is true and P ( a ) is false, it is a contradiction. If b = c , since R ( b ) is true and R ( c ) is false, it is a contradiction.

. . 3 . . . 2 . . . 1 . . . Solution. Recall Exercise (2.68) MA1100 Tutorial . . . . Tutorial Tutorial 2: Logic Let A = { 1 , 2 , . . . , 6 } and B = { 1 , 2 , . . . , 7 } . For x P A , let P ( x ) : 7 x + 4 is odd. For y P B , let Q ( y ) : 5 y + 9 is odd. Let {( ) } S = P ( x ) , Q ( y ) : x P A , y P B , P ( x ) � Q ( y ) is false . What is | S | ? For given x 0 , P ( x 0 ) is a sentence, not a truth value. P ( x ) is true for x = 1 , 3 , 5 and false for x = 2 , 4 , 6 . Q ( y ) is true for y = 2 , 4 , 6 and false for y = 1 , 3 , 5 , 7 . P ( x ) � Q ( y ) is false if P ( x ) is true and Q ( y ) is false. Thus S = { ( P ( x ) , Q ( y ) : x = 1 , 3 , 5 , y = 1 , 3 , 5 , 7 } = { P ( x ) : x = 1 , 3 , 5 } � { Q ( y ) : y = 1 , 3 , 5 , 7 } , and | S | = 3 � 4 = 12 .

. . . . . . MA1100 Tutorial Tutorial 3: Proof Schedule of Today Review concepts Tutorial: 3.10(parity), 3.18(parity, biconditional, direct proof, contrapositive), 3.20(by cases), 3.22(parity), 3.27, 3.28, 3.29, 4.4(congruent, by cases), 4.12(contrapositive, by cases), 4,15(congruent)

. . Advantage: more conditions. Disjunction in conclusion: assumption to several cases, then prove every case. Proof by cases: for convenience, we usually split the statements to get conclusion Q . Direct proof: Starting from hypothesis P , using some true Proof Theorem, lemma, corollary, proposition (need proofs). without needing a proof is called an axiom. Axiom: Basic properties that are regarded as true statement represents some object, property or other concepts. Definition: Giving the precise meaning of a word or phase that True statements True statements and Proof Review Tutorial 3: Proof MA1100 Tutorial . . . . Axioms, Definitions � Theorems, Lemmas, Propositions. Proof by contrapositive: P � Q � ( � Q ) � ( � P ) . ( P � ( Q _ R )) � (( P ^ ( � Q )) � R ) . Proving biconditionals: P � Q � ( P � Q ) ^ ( Q � P ) .

. . ab is even iff a is even or b is even. (Theorem 3.17) There are some facts: . odd Parity: n is multiplication addition Integers Review . . . . MA1100 Tutorial Tutorial 3: Proof ab is odd iff a is odd and b is odd. (Contrapositive of Theorem 3.17) Basic properties of ( Z , + , � ) : Identity: n + 0 = n and n � 1 = n ; Inverse: n + ( � n ) = 0 , but the inverse for multiplication does not exist except n = � 1 ; Commutative: n + m = m + n and m � n = n � m ; Associative: ( m + l ) + n = m + ( l + n ) and ( m � l ) � n = m � ( l � n ) ; Distributive: m � ( l + n ) = m � l + m � n , and ( m + l ) � n = m � n + l � n ; { m + n Closure: Z is closed under P Z for any m , n P Z . m � n { { 2 m + 1 even, iff there exists an integer m , such that n = 2 m odd � odd=even, odd � even=odd, even � even=even. (By definition) n is even iff n 2 is even; (Theorem 3.12) n is odd iff n 2 is odd; (Contrapositive of Theorem 3.12)

. Integers integer q . That is, for any integers a and d (here we assume that d is positive), we have the divisor, and a is called the dividend. absolute value of d . q is called the quotient, r is called the remainder, d is called . 3Ref Theorem 11.4 on page 247 Review . MA1100 Tutorial . . . Tutorial 3: Proof Divisibility: m divides n if there exists an integer q , such that n = mq . Notation: m � n , and we say that m is divisor. Negation: m does not divide n if for any integer q , n � mq . Notation: m � n . Congruence: Let a , b and n be integers with n � 1 . If n divides a � b , we say that a is congruent to b modulo n . Notation: a � b ( mod n ) . If a � b ( mod n ) , then a = b + nk for some integer k . Relation: Let a and n be integers with n � 1 . a � 0 ( mod n ) iff n � a . Division Algorithm 3 : Given two integers a and d , with d � 0 . There exist unique integers q and r such that a = qd + r and 0 � r � | d | , where | d | denotes the that a can be expressed as a = qd , a = qd + 1 , . . . , or a = qd + ( d � 1) for some For example, let d = 3 , then every integer x can be expressed as x = 3 q , x = 3 q + 1 , or x = 3 q + 2 for some integer q .

. Proof. . . . . MA1100 Tutorial Tutorial 3: Proof Review Real Numbers . { x , if x � 0; Absolute value: | x | = � x , if x � 0 . Triangle inequality: | x + y | � | x | + | y | . By definition, we have �| x | � x � | x | and �| y | � y � | y | , then � ( | x | + | y | ) � x + y � | x | + | y | , also by definition, we have | x + y | � | x | + | y | . Triangle inequality: | x � y | � | x | � | y | .

. Tutorial 3: Proof Proof. Exercise (3.10) . Tutorial MA1100 Tutorial . . . . Let x P Z . Prove that if 2 2 x is an odd integer, then 4 x is an odd integer. For any x P Z , we have 2 2 ) x = 4 x . 2 2 x = ( Therefore 4 x is odd iff 2 2 x is odd.

. Exercise (3.18) Also we can use a proof by contrapositive: assume that n is odd, we want also two methods to prove it: . There are two directions to be proved: Proof of Method 1. two methods to prove it: Tutorial . MA1100 Tutorial . . Tutorial 3: Proof . Let n P Z . Prove that ( n + 1) 2 � 1 is even if and only if n is even. “If” Suppose that n is even, we want to show that ( n + 1) 2 � 1 is even. There are First we use a direct proof: Let n = 2 k where k is an integer. Then ( n + 1) 2 � 1 = 4 k 2 + 4 k = 2(2 k 2 + 2 k ) is even. Also we can use a proof by contrapositive: assume that ( n + 1) 2 � 1 is odd, we want to show that n is odd: ( n + 1) 2 � 1 = n ( n + 2) is odd, then both n and n + 2 are odd (by contrapositive of Thm 3.17). “Only if” Suppose that ( n + 1) 2 � 1 is even, we want to show that n is even. There are First we use a direct proof: Since ( n + 1) 2 � 1 = n ( n + 2) is even, we have that n is even or n + 2 is even. If n is even, okay, there is nothing to prove; if n + 2 is even, let n + 2 = 2 k , then n = 2( k � 1) is even. to show that ( n + 1) 2 � 1 is odd: Let n = 2 m + 1 , then ( n + 1) 2 � 1 = 4 m 2 + 8 m + 3 is odd.

. . 3 . . . 2 . . . 1 . . . Proof of Method 2. Exercise (3.18) Tutorial Tutorial 3: Proof MA1100 Tutorial . . . . Let n P Z . Prove that ( n + 1) 2 � 1 is even if and only if n is even. We have ( n + 1) 2 � 1 = n 2 + 2 n + 1 � 1 = n 2 + 2 n . Since 2 n is always even, ( n + 1) 2 � 1 is even if and only if n 2 is even. n 2 is even if and only if n is even. (By theorem 3.12)

. This question is about parity, then we will split its assumptions to some cases. 2 . . . 1 . . . Proof of Method 1. . Recall MA1100 Tutorial . . . . Tutorial 3: Proof Tutorial Exercise (3.20) Prove that if n P Z , then n 3 � n is even. Let n P Z . We consider two cases. n is even. Then n = 2 a for some integer a . Thus n 3 � n = 8 a 3 � 2 a = 2(4 a 3 � a ) . Since 4 a 3 � a is an integer, n 3 � n is even. n is odd. Then n = 2 b + 1 for some integer b . Observe that n 3 � n = (2 b + 1) 3 � (2 b + 1) = 8 b 3 + 12 b 2 + 6 b + 1 � 2 b � 1 = 8 b 3 + 12 b 2 + 4 b = 2(4 b 3 + 6 b 2 + 2 b ) Since 4 b 3 + 6 b 2 + 2 b is an integer, n 3 � n is even.

. Tutorial 3: Proof Theorem 3.17); Proof of Method 2. Exercise (3.20) . Tutorial MA1100 Tutorial . . . . Theorem 3.17). Prove that if n P Z , then n 3 � n is even. Let n P Z , then n 3 � n = n ( n + 1)( n � 1) . If n is odd, then both n � 1 and n + 1 are even, therefore n 3 � n is even (By If n is even, then there is nothing to prove since even � integer=even (By

. . (also by contrapositive). a is odd and b is odd (by contrapositive). By theorem 3.17: ab is even iff a is even or b is even, we have that ab is odd iff Proof. Exercise (3.22) Tutorial Tutorial 3: Proof MA1100 Tutorial . . . . Let a , b P Z . Prove that if ab is odd, then a 2 + b 2 is even. By theorem 3.12: n 2 is even iff n is even, we have that both a 2 and b 2 are odd Since odd � odd=even, we have that a 2 + b 2 is even.

. 2 . . 1 . . . . . . . 3 . . . 4 . Which of the following is being proved? . Below is a proof of a result. . . MA1100 Tutorial Tutorial 3: Proof Tutorial Exercise (3.27) Proof: We consider two cases. . Thus a and b are even. Then a = 2 r and b = 2 s for some integers r and s . Thus a 2 � b 2 = (2 r ) 2 � (2 s ) 2 = 4 r 2 � 4 s 2 = 2(2 r 2 � 2 s 2 ) . Since 2 r 2 � 2 s 2 is an integer, a 2 � b 2 is even. a and b are odd. Then a = 2 r + 1 and b = 2 s + 1 for some integers r and s . a 2 � b 2 = (2 r + 1) 2 � (2 s + 1) 2 = (4 r 2 + 4 r + 1) � (4 s 2 + 4 s + 1) = 4 r 2 + 4 r � 4 s 2 � 4 s = 2(2 r 2 + 2 r � 2 s 2 � 2 s ) . Since 2 r 2 + 2 r � 2 s 2 � 2 s is an integer, a 2 � b 2 is even. Let a , b P Z . Then a and b are of the same parity if and only if a 2 � b 2 is even. Let a , b P Z . Then a 2 � b 2 is even. Let a , b P Z . If a and b are of the same parity, then a 2 � b 2 is even. Let a , b P Z . If a 2 � b 2 is even, then a and b are of the same parity.

. Method proved. There is only one direction, then (1) is not proved. Solution. check the proof. For these questions, we focus on the assumptions and conclusions, and no need to Tutorial . Tutorial 3: Proof MA1100 Tutorial . . . . (3) is proved. Since (4) is the converse of (3), (4) is not proved. There are two cases: a , b are both even or a , b are both odd, then (2) is not

. Tutorial 3: Proof Below is given a proof of a result. What result is being proved? . Tutorial Exercise (3.28) MA1100 Tutorial . . . . Proof: Assume that x is even. Then x = 2 a for some integer a . So 3 x 2 � 4 x � 5 = 3(2 a ) 2 � 4(2 a ) � 5 = 12 a 2 � 8 a � 5 = 2(6 a 2 � 4 a � 3) + 1 . Since 6 a 2 � 4 a � 3 is an integer, 3 x 2 � 4 x � 5 is odd. For the converse, assume that x is odd. So x = 2 b + 1 , where b P Z . Therefore, 3 x 2 � 4 x � 5 = 3(2 b + 1) 2 � 4(2 b + 1) � 5 = 3(4 b 2 + 4 b + 1) � 8 b � 4 � 5 = 12 b 2 + 4 b � 6 = 2(6 b 2 + 2 b � 3) . Since 6 b 2 + 2 b � 3 is an integer, 3 x 2 � 4 x � 5 is even.

. Tutorial 3: Proof Therefore, the result which has been proved is that for any integer x , x is even if There are two parts of the proof: . Tutorial Solution. MA1100 Tutorial . . . . The first part: x is even � 3 x 2 � 4 x � 5 is odd; The second part: x is odd � 3 x 2 � 4 x � 5 is even; This statement is equivalent to its contrapositive: x is even � 3 x 2 � 4 x � 5 is odd. and only if 3 x 2 � 4 x � 5 is odd. This can also be restated as: Let x P Z . Then x is odd if and only if 3 x 2 � 4 x � 5 is even.

. Tutorial 3: Proof Solution. Evaluate the proof of the following result. . Tutorial Exercise (3.29) MA1100 Tutorial . . . . converse. No proof has been given of the result itself. Result: Let n P Z . If 3 n � 8 is odd, then n is odd. Proof: Assume that n is odd. Then n = 2 k + 1 for some integer k . Then 3 n � 8 = 3(2 k + 1) � 8 = 6 k + 3 � 8 = 6 k � 5 = 2(3 k � 3) + 1 . Since 3 k � 3 is an integer, 3 n � 8 is odd. From “ 3 n � 8 is odd”, we want to show that “ n is odd”, but the proof shows its

. . . . . 2 . . . Using similar arguments for the remaining cases. . . 4 . 1 . 3 Tutorial . . . . MA1100 Tutorial Tutorial 3: Proof Exercise (4.4) . Proof of Method 1. we consider the following four cases. . Let x , y P Z . Prove that if 3 � x and 3 � y , then 3 � ( x 2 � y 2 ) . Assume that 3 � x and 3 � y . Then by division algorithm, we have that x = 3 p + 1 or x = 3 p + 2 for some integer p and y = 3 q + 1 or y = 3 q + 2 for some integer q . Then x = 3 p + 1 and y = 3 q + 1 . Then x 2 � y 2 = (3 p + 1) 2 � (3 q + 1) 2 = (9 p 2 + 6 p + 1) � (9 q 2 + 6 q + 1) = 3(3 p 2 + 2 p � 3 q 2 � 2 q ) . Since 3 p 2 + 2 p � 3 q 2 � 2 q is an integer, 3 � ( x 2 � y 2 ) . x = 3 p + 1 and y = 3 q + 2 . x = 3 p + 2 and y = 3 q + 1 . x = 3 p + 2 and y = 3 q + 2 .

. Tutorial 3: Proof Proof of Method 2. . Tutorial Exercise (4.4) MA1100 Tutorial . . . . Let x , y P Z . Prove that if 3 � x and 3 � y , then 3 � ( x 2 � y 2 ) . x 2 � y 2 = ( x + y )( x � y ) , then it suffices to show that at least on of ( x + y ) and ( x � y ) can be 3 divided (By result 4.3). Also consider the four cases before: x = 3 p + 1 and y = 3 q + 1 , then 3 � ( x � y ) . x = 3 p + 1 and y = 3 q + 2 , then 3 � ( x + y ) . x = 3 p + 2 and y = 3 q + 1 , then 3 � ( x + y ) . x = 3 p + 2 and y = 3 q + 2 , then 3 � ( x � y ) .

. congruent to 0 modulo 3 or neither is congruent to 0 modulo 3. Then by distributive law, left hand side is The left hand side is . . Then the If the conclusion is complicated, we may consider to show its contrapositive. The Method contrapositive is Exercise (4.12) Tutorial Tutorial 3: Proof MA1100 Tutorial . . . . Let a , b P Z . Prove that if a 2 + 2 b 2 � 0 ( mod 3) , then either a and b are both ) original statement is ( a 2 + 2 b 2 � 0) � ( ( a , b � 0) _ ( a , b � 0) (( )) ) ( � ( a 2 + 2 b 2 � 0) . � ( a , b � 0) ^ � ( a , b � 0) ( ) ( ) ( a � 0 , b � 0) _ ( a � 0 , b � 0) _ ( a � 0 , b � 0) ^ ( a � 0 , b � 0) _ ( a � 0 , b � 0) _ ( a � 0 , b � 0) . ( a � 0 , b � 0) _ ( a � 0 , b � 0) .

. Therefore, there are 2 cases, and each of which has two subcases: . . . . MA1100 Tutorial Tutorial 3: Proof Tutorial Method Then the contrapositive is . ( ) � ( a 2 + 2 b 2 � 0) . ( a � 0 , b � 0) _ ( a � 0 , b � 0)  { a = 3 p and b = 3 q + 1   a � 0 ( mod 3) and b � 0 ( mod 3)   a = 3 p and b = 3 q + 2    {   a = 3 p + 1 and b = 3 q   a � 0 ( mod 3) and b � 0 ( mod 3)    a = 3 p + 2 and b = 3 q

. . 2 . . . 1 . . . 2 . . . 1 Proof. . . . . MA1100 Tutorial Tutorial 3: Proof Tutorial . . . Case 1 Assume that a � 0 ( mod 3) and b � 0 ( mod 3) . b = 3 q + 1 . Then a 2 + 2 b 2 = (3 p ) 2 + 2(3 q + 1) 2 = 9 p 2 + 2(9 q 2 + 6 q + 1) = 9 p 2 + 18 q 2 + 12 q + 2 = 3(3 p 2 + 6 q 2 + 4 q ) + 2 . Since 3 p 2 + 6 q 2 + 4 q is an integer, 3 � ( a 2 + 2 b 2 ) and so a 2 + 2 b 2 � 0 ( mod 3) . b = 3 q + 2 . (The proof is similar to that of Subcase 1.1.) Case 2 Assume that a � 0 ( mod 3) and b � 0 ( mod 3) . a = 3 p + 1 . Then a 2 + 2 b 2 = (3 p + 1) 2 + 2(3 q ) 2 = 9 p 2 + 6 p + 1 + 18 q 2 = 9 p 2 + 6 p + 18 q 2 + 1 = 3(3 p 2 + 2 p + 6 q 2 ) + 1 . Since 3 p 2 + 2 p + 6 q 2 is an integer, 3 � ( a 2 + 2 b 2 ) and so a 2 + 2 b 2 � 0 ( mod 3) . a = 3 p + 2 . (The proof of each subcase is similar to that of Subcase 2.1.)

. Exercise (4.12) statement is There is another method based on Disjunction in conclusion: Method congruent to 0 modulo 3 or neither is congruent to 0 modulo 3. . then it has been done. Tutorial Tutorial 3: Proof MA1100 Tutorial . . . . Let a , b P Z . Prove that if a 2 + 2 b 2 � 0 ( mod 3) , then either a and b are both ( P � ( Q _ R )) � (( P ^ ( � Q )) � R ) . If we denote P : a 2 + 2 b 2 � 0 , Q : a � 0 , b � 0 , R : a � 0 , b � 0 , then the original P � ( Q _ R ) . By Disjunction in conclusion: if we prove that P ^ ( � Q ) � R , i.e. a 2 + 2 b 2 � 0 , a � 0 or b � 0 � a � 0 , b � 0 ,

. Proof. . . . . MA1100 Tutorial Tutorial 3: Proof Tutorial Exercise (4.15) . Observe that Let a , b P Z . Show that if a � 5 ( mod 6) and b � 3 ( mod 4) , then 4 a + 6 b � 6 ( mod 8) . Assume that a � 5 ( mod 6) and b � 3 ( mod 4) . Then 6 � ( a � 5) and 4 � ( b � 3) . Thus a � 5 = 6 x and b � 3 = 4 y , where x , y P Z . So a = 6 x + 5 and b = 4 y + 3 . 4 a + 6 b = 4(6 x + 5) + 6(4 y + 3) = 24 x + 20 + 24 y + 18 = 24 x + 24 y + 38 = 8(3 x + 3 y + 4) + 6 . Since 3 x + 3 y + 4 is an integer, 8 � (4 a + 6 b � 6) and so 4 a + 6 b � 6 ( mod 8) .

. . . . . . MA1100 Tutorial Tutorial 4: Proof Schedule of Today Review concepts Tutorial: 4.23(absolute value), 4.33(set relation), 4.42(set contradiction), 5.32(existence, proof by contradiction), 5.33(existence, intermediate value theorem, uniqueness) relation), 5.4(parity, disprove), 5.13(rational, irrational, proof by contradiction), 5.22(parity, existence, proof by

. . For direct proof, we have one assumption P ; When it is easy to work with the negation. “ p is an irrational number”. When there is no direct proof: “there do not exist...”, “ A is an emptyset”, When to use: to get a contradiction. contradiction; 4Compare with disjunction in conclusion. Direct Proof, Proof by Contrapositive, Proof by contradiction; Proof by Contradiction Review Tutorial 4: Proof MA1100 Tutorial . . . . Prove R is true: Assume � R is true, try to get a contradiction; Prove ( @ x ) R ( x ) is true: Assume ( D x ) � R ( x ) is true, try to get a Prove ( @ x ) P ( x ) � Q ( x ) is true: Assume ( D x ) P ( x ) ^ ( � Q ( x )) is true, try Advantage: For implication P � Q , we have more assumption 4 to work with: For proof by contradiction, we have more assumption � Q .

. . 2 Make arguments why such objects have to exist; 1 Use when specific examples are not easy or not possible to find; Non-constructive proof: 2 Justify that the given examples satisfy the stated conditions. 1 Give a specific example of such objects; Constructive proof: Two approaches: 3 Use definitions, axioms or results that involves existence statements. Three types: Existence Proof Review Tutorial 4: Proof MA1100 Tutorial . . . . ( D x ) P ( x ) : ( D x )( @ y ) P ( x , y ) : ( @ x )( D y ) P ( x , y ) :

. Irrational An irrational number is a real number that is not a rational number. no common factor which is greater than 1. This property is very . Irrational Rational Rational Decimal Decimal . Properties useful for proof by contradiction. m MA1100 Tutorial Irrational Numbers Review Tutorial 4: Proof . . . . Rational A rational number is a real number that can be written as a quotient n where m and n are integers, with n � 0 . Rational � Rational = Rational , Rational � Irrational = Irrational , Irrational � Irrational = ?(it depends) .  finite decimal ,   non-terminating recurring decimal ,   non-terminating non-recurring decimal , n with n � 0 is in lowest term if m and n have Lowest Term A rational number m

. A c Logic Meaning Set Algebra of Sets: 2 Show that the element satisfies the given property. 1 Choose an arbitrary element; Element-Chasing Method: Sets Relations Review Tutorial 4: Proof MA1100 Tutorial . . . . . 3 Using this method, we can prove: A � B , A � B , A = B , A � B . A X B x P A and x P B P ^ Q Set Operations: P : x P A , Q : x P B , A Y B x P A or x P B P _ Q x R A � P A � B a P A and x R B P ^ ( � Q ) Idempotent A X A = A , A Y A = A ; Commutative A X B = B X A , A Y B = B Y A ; Associative ( A X B ) X C = A X ( B X C ) , ( A Y B ) Y C = A Y ( B Y C ) ; Distributive A X ( B Y C ) = ( A X B ) Y ( A X C ) , A Y ( B X C ) = ( A Y B ) X ( A Y C ) ; Double Complement ( A c ) c = A ; De Morgan ( A X B ) c = A c Y B c , ( A Y B ) c = A c X B c ;

. Exercise (4.23) Proof. . Recall For these questions, based on definition of absolute value, we consider some cases. . Method Tutorial Tutorial 4: Proof MA1100 Tutorial . . . . Let x , y P R . Prove that | xy | = | x | � | y | .  if x � 0; { { x ,   x , if x � 0; x , if x � 0; Equivalent definitions: | x | = if x � 0 . = if x � 0 . = 0 , if x = 0; � x , � x ,   � x , if x � 0 . Assume x � 0 and y � 0 : then xy � 0 , and | xy | = xy , | x | = x , | y | = y . Therefore, | xy | = xy = | x | � | y | . Assume x � 0 and y � 0 : then xy � 0 , and | xy | = � xy , | x | = x , | y | = � y . Therefore, | xy | = � xy = | x | � | y | . Assume x � 0 and y � 0 : then xy � 0 , and | xy | = � xy , | x | = � x , | y | = y . Therefore, | xy | = � xy = | x | � | y | . Assume x � 0 and y � 0 : then xy � 0 , and | xy | = xy , | x | = � x , | y | = � y . Therefore, | xy | = xy = ( � x )( � y ) = | x | � | y | .

. Direct Proof: . . . . MA1100 Tutorial Tutorial 4: Proof Tutorial Exercise (4.33) Proof. There are two directions: . Let A and B be sets. Prove that A Y B = A X B if and only if A = B . “If” Assume A = B , then A Y B = A = A X B . “Only if” Assume A Y B = A X B , we want to show A = B . Since A = B iff A � B and B � A , it suffices to show A � B and B � A . Using element-chasing method: for any x P A , then x P A Y B = A X B , therefore x P B , that is, A � B ; By symmetry, we also get B � A . Proof by contrapositive: assume A � B , that is A � B or B � A , we want to show that A Y B � A X B : When A � B : By definition, there exists a P A and a R B . Since a R B , we have a R A X B . On the other hand, a P A implies that a P A Y B . Therefore A Y B � A X B . When B � A : Similar.

. try to add more conditions.) . sufficient condition for original statement.) . . . 2 . . . . 3 . . . 4 . 1 . Solution. . . . . MA1100 Tutorial Tutorial 4: Proof Tutorial Exercise (4.42) For sets A and B , fine a necessary and sufficient condition for Recall Method Start from assumption, we try to get some necessary(sufficient) condition, then try to prove the necessary(sufficient) condition is sufficient(necessary). ( A � B ) X ( B � A ) = H . Verify that this condition is necessary and sufficient. Cartesian Product: A � B = { ( x , y ) : x P A , y P B } . Assume ( A � B ) X ( B � A ) � H , then there exists ( x , y ) P ( A � B ) X ( B � A ) , that is x P A , y P B and x P B , y P A , i.e. A X B � H . ( A X B = H is Guess: when A X B � H , we have ( A � B ) X ( B � A ) � H . (If not, we must Assume A X B � H , then there exists x P A X B , therefore ( x , x ) P ( A � B ) X ( B � A ) , i.e. ( A � B ) X ( B � A ) � H . Therefore, we find a necessary and sufficient condition: A X B = H .

. odd . odd 1 3 6 10 15 21 28 36 odd odd even even odd 7 21 even even odd 45 36 28 15 even 10 6 3 even even even 8 6 odd For these questions, generally there are two methods: . . . . MA1100 Tutorial Tutorial 4: Proof Tutorial Exercise (5.4) is odd. Method When disproving a statement, we only need a counterexample, and one 5 counterexample is enough. odd is even. enumerating method to find the number. Solution of Method 1. 4 3 2 1 n Start from assumption, to get some properties, then using the properties and Disprove the statement: Let n P N . If n ( n +1) is odd, then ( n +1)( n +2) 2 2 Enumerating: try 1 , 2 , 3 , . . . ; We want to find an integer n , such that n ( n +1) is odd and ( n +1)( n +2) 2 2 n ( n +1) 2 ( n +1)( n +2) 2 Thus n = 2 is a counterexample.

. . . . 1 . is odd, then is even, We can use enumerating method for the restricted set . of n . . 2 4 . . . 3 . . . is even, we want to find some properties is odd. . . . . MA1100 Tutorial Tutorial 4: Proof Tutorial Exercise (5.4) . Solution of Method 2. Disprove the statement: Let n P N . If n ( n +1) is odd, then ( n +1)( n +2) 2 2 Assume that n ( n +1) is odd and ( n +1)( n +2) 2 2 Since ( n +1)( n +2) = n ( n +1) + ( n + 1) , n ( n +1) is odd and ( n +1)( n +2) 2 2 2 2 we have n is even, say 2 k , k P N ; Substitute n = 2 k to n ( n +1) and ( n +1)( n +2) : k (2 k + 1) is odd and 2 2 ( k + 1)(2 k + 1) is even, therefore k is odd, say 2 p + 1 , p � 0 ; We get n = 2(2 p + 1) , p � 0 . { n P N : n = 2(2 p + 1) , p � 0 } . Letting p = 0 , we get that n ( n +1) 2 ( n +1)( n +2) is even when n = 2 . 2

. . . . 1 Assume that there exist an irrational number a and a nonzero rational number b such that ab is rational. . . Proof. 2 . . . 3 p . . For these questions, in general, we use a proof by contradiction. Tutorial 4: Proof . . . . MA1100 Tutorial Method number, which is a contradiction. Tutorial Exercise (5.13V) Prove that the product of an irrational number and a nonzero rational number is irrational. Recall R has two parts: rational numbers Q and irrational number R � Q . Any rational q , where p , q P Z and q � 0 . number a can be expressed as p s , where r , s P Z and r , s � 0 ; ab can be By definition, b can be expressed as r q , where p , q P Z and q � 0 . expressed as p Then a = bq = sp rq . Since sp , rq P Z and rq � 0 , it follows that a is a rational

. Proof. or of opposite parity. here we consider two cases, according to whether x and y are of the same parity Generally we may consider four cases based on the parities of x and y . While, 2 . . . 1 . . . Here we use a proof by contradiction: . infinite. For these questions, it is difficult to give a direct proof, since the number of integers is . . . . MA1100 Tutorial Tutorial 4: Proof Tutorial Exercise (5.22V) Method Let m be a positive integer of the form m = 2 s , where s is an odd integer. Prove that there do not exist positive integers x and y such that x 2 � y 2 = m . Assume that there exist positive integers x and y such that x 2 � y 2 = m = 2 s . Then ( x + y )( x � y ) = 2 s , where s is an odd integer. If x and y are of the same parity, then both x + y and x � y are even, then 2 s has factor 4, i.e. s has factor 2, it is a contradiction; If x and y are of opposite parity, then both x + y and x � y are odd, therefore 2 s is odd. Contradiction.

. . . . . 1 Assume that there exist nonzero real numbers a and b such that . . Proof. 2 Raising both sides to the 6th power, we obtain Thus . . . 3 Here we use a proof by contradiction: . numbers is infinite. For these questions, it is difficult to give a direct proof, since the number of real . . . . MA1100 Tutorial Tutorial 4: Proof Tutorial Exercise (5.32V) Show that there exist no nonzero real numbers a and b such that 5Non-negative+Non-negative=0 iff each of them is zero. Method √ √ a 2 + b 2 = 3 a 3 + b 3 . √ √ a 2 + b 2 = a 3 + b 3 . 3 a 6 + 3 a 4 b 2 + 3 a 2 b 4 + b 6 = a 6 + 2 a 3 b 3 + b 6 . 3 a 2 � 2 ab + 3 b 2 = ( a � b ) 2 + 2 a 2 + 2 b 2 = 0 . Since this can only occur when a = b = 0 5 , we have a contradiction.

. For existence of these questions, we will use Intermediate Value Theorem if we can not Since f is a polynomial function, it is continuous on the set of all real numbers 1 . . . . Proof of existence. 2 . Recall solve the equation directly. Method . Exercise (5.33) Tutorial Tutorial 4: Proof MA1100 Tutorial . . . . . Prove that there exist a unique real number solution to the equation x 3 + x 2 � 1 = 0 between x = 2 3 and x = 1 . Intermediate Value Theorem: If the function y = f ( x ) is continuous on the interval [ a , b ] , and f ( a ) f ( b ) � 0 , then there is a c P [ a , b ] such that f ( c ) = 0 . Let f ( x ) = x 3 + x 2 � 1 . and so f is continuous on the interval [2/3 , 1] . Because f (2/3) = � 7/27 � 0 and f (1) = 1 � 0 , it follows by the Intermediate Value Theorem that there is a number c between x = 2/3 and x = 1 such that f ( c ) = 0 . Hence c is a solution.

. . . 3 . 2 . . . 1 . . . . MA1100 Tutorial . . . Proof of uniqueness. . Tutorial 4: Proof Tutorial Method For uniqueness, in general, let c 1 and c 2 be the numbers each of which satisfies the condition, and then prove c 1 = c 2 . We now show that c is the unique solution of f ( x ) = 0 between 2/3 and 1. Let c 1 and c 2 be solutions of f ( x ) = 0 between 2/3 and 1. Then c 3 1 + c 2 1 � 1 = 0 and c 3 2 + c 2 2 � 1 = 0 . Hence c 3 1 + c 2 1 � 1 = c 3 2 + c 2 2 � 1 , implying that c 3 1 + c 2 1 = c 3 2 + c 2 2 and so c 3 1 � c 3 2 + c 2 1 � c 2 2 = ( c 1 � c 2 )( c 2 1 + c 1 c 2 + c 2 2 ) + ( c 1 � c 2 )( c 1 + c 2 ) = ( c 1 � c 2 )( c 2 1 + c 1 c 2 + c 2 2 + c 1 + c 2 ) = 0 . Since c 2 1 + c 1 c 2 + c 2 2 + c 1 + c 2 � 0 (because 2/3 � c 1 , c 2 � 1 ), we obtain c 1 � c 2 = 0 and so c 1 = c 2 .

. Close book with 1 helpsheet; Results without proof: http://wiki.nus.edu.sg/display/MA1100/MA1100+Home Wiki for MA1100: S9a-02-03. Oct 1st, Thursday, 09:00–11:00, 13:00–17:00, 19:00–21:00; Sept 28th, Monday, 13:00–17:00; Consultation: Venue: MPSH1; . Time: Oct 2nd, 12-2pm; Mid-Term Exam MA1100 Tutorial . . . . http://wiki.nus.edu.sg/display/MA1100/MA1100+Basic+Results

. . . . . . MA1100 Tutorial Tutorial 5: Mathematical Induction Schedule of Today Review concepts Tutorial: 6.8, 6.9, 6.16, 6.20, 6.22, 6.35, 6.37, 6.51

. smallest element. . . . . MA1100 Tutorial Tutorial 5: Mathematical Induction Review Principle of Mathematical Induction . true. Axiom of Induction If T is a subset of N , such that: 1 P T ; For every k P N , if k P T , then k + 1 P T . Then T = N . Well-Ordered Let H � S � R , S is well-ordered if every nonempty subset of S has Well-Ordering Principle The set N is well-ordered. PMI Let P ( n ) be an open sentence, such that If P (1) is true; For all k P N , if P ( k ) is true, then P ( k + 1) is true. Then P ( n ) is true for all n P N . SPMI Let P ( n ) be an open sentence, such that If P (1) is true; For all k P N , if P (1) , P (2) , . . . , P ( k ) are true, then P ( k + 1) is Then P ( n ) is true for all n P N .

. Summary the conclusion you get. . . . . MA1100 Tutorial Tutorial 5: Mathematical Induction Review Applications and Generalizations . where M is an integer; Generalization of universal set with some “good” order. PMI(or SPMI) can be used to prove ( @ n P N ) P ( n ) is true: Identify universal set and open sentence P ( n ) ; Base case: prove P (1) is true; Inductive step: for all k P N , if P ( k ) (or P (1) ^ P (2) ^ � � � ^ P ( k ) ) is true, prove P ( k + 1) is true; { n : n P Z , n � M } with the order: P ( M ) � P ( M +1) � P ( M +2) � � � � � P ( M + n ) � P ( M + n +1) � � � � Z with the order: � P ( � n ) � � � � � P ( � 2) � P ( � 1) � P (0) � P (1) � P (2) � � � � � P ( n ) � � � � ; Q + with order in lecture notes; { n P N : n = 3 p + 1 , p P N } with the natural order: P (1) � P (4) � P (7) � � � � � P (3 p + 1) � P (3 p + 4) � � � � .

. for . . . . MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial Exercise (6.8) . Solution for (a). represent geometrically? every positive integer n . (a) We have seen that 1 2 + 2 2 + � � � + n 2 is the number of the squares in an n � n square composed of n 2 1 � 1 squares. What does 1 3 + 2 3 + 3 3 + � � � + n 3 (b) Use mathematical induction to prove that 1 3 + 2 3 + 3 3 + � � � + n 3 = n 2 ( n +1) 2 4 Let C be an n � n � n cube composed of n 3 1 � 1 � 1 cubes. Then the number of different cubes that C contains is 1 3 + 2 3 + 3 3 + � � � + n 3 .

. Proof for (b). (3) By the Principle of Mathematical Induction, we have : . ; . for every positive integer n . Tutorial . MA1100 Tutorial . . . Tutorial 5: Mathematical Induction Universal set { n : n P N } , and P ( n ) : 1 3 + 2 3 + � � � + n 3 = n 2 ( n +1) 2 4 (1) Base case: when n = 1 , LHS = 1 3 = 1 = 1 2 (1+1) 2 = RHS, i.e. P (1) is true; 4 (2) Inductive step: for all k � 1 : Assume that P ( k ) is true, i.e. 1 3 + 2 3 + � � � + k 3 = k 2 ( k +1) 2 4 Then we want to show that P ( k + 1) is true, i.e. 1 3 + 2 3 + � � � + ( k + 1) 3 = ( k +1) 2 ( k +1+1) 2 4 1 3 + 2 3 + � � � + ( k + 1) 3 = 1 3 + 2 3 + � � � + k 3 +( k + 1) 3 � �� � k 2 ( k +1) 2 /4 + ( k + 1) 3 = ( k + 1) 2 ( k 2 + 4 k + 4) = k 2 ( k + 1) 2 4 4 = ( k + 1) 2 ( k + 1 + 1) 2 4 1 3 + 2 3 + 3 3 + � � � + n 3 = n 2 ( n +1) 2 4

. for every positive (3) By the Principle of Mathematical Induction, we have : ; . . Proof. integer n . for every positive integer n . Exercise (6.9) Tutorial Tutorial 5: Mathematical Induction MA1100 Tutorial . . . . Prove that 1 � 3 + 2 � 4 + 3 � 5 + � � � + n ( n + 2) = n ( n +1)(2 n +7) 6 Universal set is { n : n P N } , and P ( n ) : 1 � 3 + 2 � 4 + � � � + n ( n + 2) = n ( n +1)(2 n +7) 6 (1) Base case: when n = 1 , LHS = 1(1 + 2) = 3 = 1 � 2 � 9 = RHS, i.e. P (1) is true; 6 (2) Inductive step: for all k � 1 : Assume that P ( k ) is true, i.e. 1 � 3 + 2 � 4 + � � � + k ( k + 2) = k ( k +1)(2 k +7) 6 Then we want to show that P ( k + 1) is true, i.e. 1 � 3 + 2 � 4 + � � � + ( k + 1)( k + 1 + 2) = ( k +1)( k +1+1)(2( k +1)+7) 6 1 � 3 + 2 � 4 + � � � + ( k + 1)( k + 1 + 2) = 1 � 3 + 2 � 4 + � � � + k ( k + 2) +( k + 1)( k + 1 + 2) � �� � k ( k +1)(2 k +7)/6 = ( k + 1)( k + 1 + 1)(2( k + 1) + 7) 6 1 � 3 + 2 � 4 + 3 � 5 + � � � + n ( n + 2) = n ( n +1)(2 n +7) 6

. Tutorial k . n . Proof. Exercise (6.16,T) every positive integer n . Tutorial 5: Mathematical Induction . . . MA1100 Tutorial . Prove that 1 + 1 4 + 1 1 n 2 � 2 � 1 9 + � � � + n for every positive integer n . Universal set is { n : n P N } , and P ( n ) : 1 + 1 4 + 1 1 n 2 � 2 � 1 9 + � � � + (1) Base case: when n = 1 , LHS = 1 = 2 � 1 = RHS, i.e. P (1) is true; (2) Inductive step: for all k � 1 : Assume that P ( k ) is true, i.e. 1 + 1 4 + 1 1 k 2 � 2 � 1 9 + � � � + k ; Then we want to show 1 + 1 4 + 1 1 1 9 + � � � + ( k +1) 2 � 2 � ( k +1) : 1 + 1 ( k + 1) 2 = 1 + 1 1 4 + � � � + 1 1 4 + � � � + + k 2 ( k + 1) 2 � �� � � 2 � 1 ( k + 1) 2 = 2 � k 2 + k + 1 � 2 � 1 1 k + k ( k + 1) 2 k 2 + k 1 � 2 � k ( k + 1) 2 = 2 � k + 1 (3) By the Principle of Mathematical Induction, 1 + 1 4 + 1 1 n 2 � 2 � 1 9 + � � � + n for

. nonnegative integer n . Remark proof: . Proof. Exercise (6.20,T) Tutorial Tutorial 5: Mathematical Induction MA1100 Tutorial . . . . some integer a ; Prove that 7 � (3 2 n � 2 n ) for every nonnegative integer n . Universal set is { n P Z : n � 0 } , and P ( n ) : 7 � (3 2 n � 2 n ) . (1) Base case: when n = 0 , 3 0 � 2 0 = 0 and 7 � 0 , i.e. P (0) is true; (2) Inductive step: for all k � 0 : Assume that P ( k ) is true, that is, 7 � (3 2 k � 2 k ) , i.e. 3 2 k = 2 k + 7 a for Then we want to show 7 � (3 2( k +1) � 2 k +1 ) , and it suffices to show that 3 2( k +1) � 2 k +1 has factor 7: 3 2( k +1) � 2 k +1 = 3 2 � 3 2 k � 2 � 2 k = 9(2 k + 7 a ) � 2 � 2 k = 7(2 k + 9 a ) . (3) By the Principle of Mathematical Induction, we have 7 � (3 2 n � 2 n ) for every When n � 1 , by a n � b n = ( a � b )( a n � 1 + a n � 2 b + � � � + b n � 1 ) , we have a direct 3 2 n � 2 n = 9 n � 2 n = (9 � 2) c = 7 c , where c = 9 n � 1 2 + 9 n � 2 2 2 + � � � + 92 n � 1 is an integer.

. Exercise (6.22,T) Hypothesis Associated law (3) By the Principle of Mathematical Induction, we have . Associated law Proof. Tutorial Tutorial 5: Mathematical Induction MA1100 Tutorial . . . . De Morgan’s law Prove that if A 1 , A 2 , . . . , A n are any n � 2 sets, then A 1 X � � � X A n = A 1 Y � � � Y A n . Universal set is { n P Z : n � 2 } , and P ( n ) : A 1 X A 2 X � � � X A n = A 1 Y A 2 Y � � � Y A n . (1) Base case: when n = 2 , by De Morgan’s law, we have A 1 X A 2 = A 1 Y A 2 for any sets A 1 , A 2 , i.e. P (2) is true; (2) Inductive step: for all k � 2 : Assume that P ( k ) is true, i.e. A 1 X � � � X A k = A 1 Y � � � Y A k for any k sets A 1 , . . . , A k ; Then we want to show that P ( k + 1) is true, i.e. A 1 X � � � X A k +1 = A 1 Y � � � Y A k +1 for any k + 1 sets A 1 , . . . , A k +1 : A 1 X A 2 X � � � X A k +1 = ( A 1 X � � � X A k ) X A k +1 = A 1 X � � � X A k Y A k +1 ( ) = A 1 Y � � � Y A k Y A k +1 = A 1 Y � � � Y A k +1 A 1 X � � � X A n = A 1 Y � � � Y A n for any n sets.

. Tutorial Solution for (a). (a) Define the sequence of Fibonacci numbers by means of a recurrence relation. called Fibonacci numbers. . Exercise (6.35) Tutorial 5: Mathematical Induction MA1100 Tutorial . . . . Consider the sequence F 1 , F 2 , F 3 , . . . , where F 1 = 1 , F 2 = 1 , F 3 = 2 , F 4 = 3 , F 5 = 5 , and F 6 = 8 . The terms of this sequence are (b) Prove that 2 � F n if and only if 3 � n . The sequence { F n } is defined recursively by F 1 = 1 , F 2 = 1 , and F n = F n � 1 + F n � 2 for n � 3 .

. recurrent relation, we have . . . . MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial Proof for (b) “if” case. . If 3 � n , we want to show 2 � F n . Universal set is { n : n = 3 q , q P N } = { 3 , 6 , . . . , 3 k , 3( k + 1) , . . . } , and P ( n ) : 2 � F n . (1) Base case: when n = 3 and F 3 = 2 , so 2 � F 3 , i.e. P (3) is true. (2) Inductive step, for all n of form n = 3 k with k P N : Assume that P (3 k ) is true, i.e. 2 � F 3 k ; Then we want to show that P (3( k + 1)) is true, i.e. 2 � F 3( k +1) : By F 3( k +1) = F 3 k +2 + F 3 k +1 = F 3 k +1 + F 3 k + F 3 k +1 = 2 F 3 k +1 + F 3 k which is even, i.e. 2 � F 3( k +1) . (3) By the Principle of Mathematical Induction, we have that 2 � F n for all n which satisfies 3 � n .

. . . . . . MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial Proof for (b) “only if” case. If 3 � n , we want to show 2 � F n . Using prove by cases: n = 3 q + 1 for q � 0 . Universal set is { n : n = 3 q + 1 , q � 0 , q P Z } = { 1 , 4 , 7 , . . . , 3 k + 1 , 3( k + 1) + 1 , . . . } , Q ( n ) : 2 � F n . (1) Base case: when n = 1 , F 1 = 1 , so 2 � F 1 , i.e. Q (1) is true. (2) Inductive step, for all n of form n = 3 k + 1 for k � 0 : Assume that Q (3 k + 1) is true, i.e. 2 � F 3 k +1 ; Then we want to show that Q (3( k + 1) + 1) is also true, i.e. 2 � F 3( k +1)+1 : By recurrent relation, we have F 3( k +1)+1 = F 3 k +3 + F 3 k +2 = F 3 k +2 + F 3 k +1 + F 3 k +2 = 2 F 3 k +2 + F 3 k +1 which is odd, i.e. 2 � F 3( k +1)+1 . (3) By the Principle of Mathematical Induction, we have that 2 � F n for all n which is of form n = 3 q + 1 . n = 3 q + 2 for q � 0 . Universal set is { n : n = 3 q + 2 , q � 0 , q P Z } = { 2 , 5 , 8 , . . . , 3 k + 2 , 3( k + 1) + 2 , . . . } , R ( n ) : 2 � F n . By similar method, we can prove 2 � F n for all n of form n = 3 q + 2 .

. . . . . . MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial Exercise (6.37) Use the Strong Principle of Mathematical Induction to prove that for each integer Proof. n � 12 , there are nonnegative integers a and b such that n = 3 a + 7 b . Universal set is { n P Z : n � 12 } , and P ( n ) : n = 3 a + 7 b for some integers a , b � 0 . (1) Base case: for n = 12 , it is clear that 12 = 3 � 4 , that is, we can choose a = 4 and b = 0 , such that 12 = 3 a + 7 b , i.e. P (12) holds; (2) Inductive step: for all k � 12 : Assume that for every integer i with 12 � i � k , there exist integers a , b � 0 such that i = 3 a + 7 b ; We want to show that there exist integers x , y � 0 such that k + 1 = 3 x + 7 y ; Since 13 = 3 � 2 + 7 � 1 and 14 = 3 � 0 + 7 � 2 , we may assume that k � 14 ; Since k � k � 2 � 12 , there exist integers c , d � 0 such that k � 2 = 3 c + 7 d ; Hence k + 1 = 3( c + 1) + 7 d . (3) By the Strong Principle of Mathematical Induction, for each integer n � 12 , there are integers a , b � 0 such that n = 3 a + 7 b .

. 5 1 . . . . 2 (3) By Principle of Mathematical Induction, the sum of the interior angles of an . . . . . 3 . . . . . . . By an n -gon, we mean an n -sided polygon. So a 3-gon is a triangle and a 4-gon is a . . . . MA1100 Tutorial Tutorial 5: Mathematical Induction Tutorial Exercise (6.51) 4 Proof. angles of an n -gon. quadrilateral. It is well known that the sum of the interior angles of a triangle is 180 0 . Use induction to prove that for every integer n � 3 , the sum of the interior angles of an n -gon is ( n � 2) � 180 0 . For convenience, we use some notations: π = 180 0 , S n denotes the sum of the interior Universal set is { n P Z : n � 3 } , and P ( n ) : S n = ( n � 2) π . (1) Base case: for n = 3 , S 3 = π = (3 � 2) � 180 0 , i.e. P (3) holds; (2) Inductive step: for all k � 3 , Let Q k +1 be a ( k + 1) -gon whose vertices are v 1 , v 2 , . . . , v k , v k +1 and whose edges are v 1 v 2 , v 2 v 3 , . . . , v k v k +1 , v k +1 v 1 ; Then let Q k be the k -gon such that whose vertices are v 1 , v 2 , . . . , v k and whose edges are v 1 v 2 , v 2 v 3 , . . . , v k v 1 , Q 3 be the 3-gon whose vertices are v k , v k +1 , v 1 and whose edges are v k v k +1 , v k +1 v 1 , v 1 v k ; Observe that S k +1 = S k + S 3 ; By the induction hypothesis, S k is ( k � 2) π and S 3 is π ; Therefore, S k +1 is ( k � 2) π + π = ( k � 1) π . n -gon is ( n � 2) � 180 0 .

. . . . . . MA1100 Tutorial Tutorial 6: Relations Schedule of Today Review concepts Tutorial: 8.6, 8.9, 8.10, 8.14, 8.17, 8.20, 8.23, 8.26

. . Let R be an equivalence relation on A , then the collection C of all equivalence class of n determined by the relation R . symmetric, transitive relation on A . Let R be a relation on A . R is an equivalence relation if it is a reflexive, Let R be a relation on A : classes determined by R is a partition of the set A . Review . Tutorial 6: Relations . MA1100 Tutorial . . Let A , B be sets. A relation R from A to B is a subset of A � B , i.e. R = { ( a , b ) P A � B : condition on a and b } . If ( x , y ) P R , then x is related to y . Notation: ( x , y ) P R , x � y , x � R y , xRy . Let A be a set. A relation R on A is the subset of A � A , i.e. R = { ( a , b ) P A � A : condition on a and b } . R is reflexive on A if “for all x P A , xRx ”; R is symmetric on A if “for all x , y P A , if xRy , then yRx ”; R is transitive on A if “for all x , y , z P A , if xRy and yRz , then xRz ”. Let R be an equivalence relation on A . For each n P A , let [ n ] R = { x P A : ( x , n ) P R } = { x P A : ( n , x ) P R } . We call this an equivalence

. Symmetric . . . . MA1100 Tutorial Tutorial 6: Relations Review Relation Relation from A to B Relation on a Set A . Reflexive when B = A ❄ ✲ Transitive ✲ Equiv Relation R ✛ Equiv Classes { [ n ] R : n P A } ✲ ✻ Thm 8 . 3 Thm 8 . 4 ❄ ✛ Partition { [ n ] R : n P A }

. Tutorial Solution. your answers. properties reflexive, symmetric, and transitive does the relation R possess? Justify . Exercise (8.6) transitive. Tutorial 6: Relations MA1100 Tutorial . . . . Let S = { a , b , c } . Then R = { ( a , a ) , ( a , b ) , ( a , c ) } is a relation on S . Which of the Not reflexive: ( b , b ) R R , ( c , c ) R R ; Not symmetric: ( a , b ) P R but ( b , a ) R R , ( a , c ) P R but ( c , a ) R R ; Transitive: The only ordered pairs ( x , y ) and ( y , z ) that belong to R are where ( x , y ) = ( a , a ) , since y has only one choice a . The possible choices for ( y , z ) in R are ( a , a ) , ( a , b ) , and ( a , c ) . In every case, ( x , z ) = ( y , z ) P R and so R is

. Tutorial 6: Relations Solution. symmetric, and transitive does the relation R possess? Justify your answers. . Tutorial Exercise (8.9T) MA1100 Tutorial . . . . A relation R is defined on Z by aRb if | a � b | � 2 . Which of the properties reflexive, R = { ( a , b ) : a , b P Z , | a � b | � 2 } . Reflexive: for all n P Z , since | n � n | = 0 � 2 , we have nRn ; Symmetric: for all m , n P Z such that mRn , i.e | m � n | � 2 , it can be rewritten as | n � m | � 2 , then we have nRm ; Not transitive: (0 , 2) P R , and (2 , 4) P R , but (0 , 4) R R .

. Tutorial We focus on a : Solution. . Exercise (8.10) way for partition, that is, there is only one equivalence relation. Tutorial 6: Relations MA1100 Tutorial . . . . Let A = { a , b , c , d } . How many relations defined on A are reflexive, symmetric, and transitive and contain the ordered pairs ( a , b ) , ( b , c ) , ( c , d ) ? Since R is reflexive, we have ( a , a ) P R ; By the assumption, we have ( a , b ) P R ; By the assumption ( a , b ) , ( b , c ) P R , since R is transitive, we have ( a , c ) P R ; Similarly, we have ( a , d ) P R . Then [ a ] R = { a , b , c , d } = A . So the associated partition is { A } , and there is only one

. Tutorial Solution. determine these equivalence classes and determine all elements of R . . Exercise (8.14) and bRf . If there are three distinct equivalence classes resulting from R , then Tutorial 6: Relations MA1100 Tutorial . . . . contradictions easily. Let R be an equivalence relation on A = { a , b , c , d , e , f , g } such that aRc , cRd , dRg , R is an equivalence relation and aRc , cRd , dRg : a , c , d , g is in the same equivalence class, namely [ a ] R . Moreover, | [ a ] R | � 4 ; Similarly, b , f is in the same equivalence class, namely [ b ] R . Moreover, | [ b ] R | � 2 ; Similarly, e is in the equivalence class [ e ] R . Moreover, | [ e ] R | � 1 ; Notice: Now we do not know whether [ a ] R , [ b ] R , [ e ] R are pairwise disjoint; Since | A | = 7 , and there are three distinct equivalence classes, therefore [ a ] R , [ b ] R , [ e ] R are pairwise disjoint and the three distinct equivalence classes are just [ a ] R = { a , c , d , g } , [ b ] R = { b , f } , [ e ] R = { e } . Otherwise, you can find

. Solution. . . . . MA1100 Tutorial Tutorial 6: Relations Tutorial Exercise (8.17T) . is an equivalence relation on A . Determine the distinct equivalence classes. Let A = { 1 , 2 , 3 , 4 , 5 , 6 } . The relation R = { (1 , 1) , (1 , 5) , (2 , 2) , (2 , 3) , (2 , 6) , (3 , 2) , (3 , 3) , (3 , 6) , (4 , 4) , (5 , 1) , (5 , 5) , (6 , 2) , (6 , 3) , (6 , 6) } The elements each of which has relation with 1 are 1 and 5 , then the equivalence class of 1 is [1] R = { 1 , 5 } = [5] R ; The elements each of which has relation with 2 are 2 , 3 and 6 , then the equivalence class of 2 is [2] R = { 2 , 3 , 6 } = [3] R = [6] R ; The element which has relation with 4 is just 4 itself, then the equivalence class of 4 is [4] R = { 4 } . Therefore there are three distinct equivalence classes, namely { 1 , 5 } , { 2 , 3 , 6 } and { 4 } .